The complete method for decomposing fractions is obvious . For example $y = \frac{2x+1}{(x-1)(x+3)} = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)} \Rightarrow $$ \left\{ \begin{array}{c} a+b= 2 \\ 3a - b = 1 \end{array} \right. $$ \Rightarrow a = 3/4 , b= 5/4 $
But I was interested in finding a quick way . So I did another operation : $y = \frac{2x+1}{(x-1)(x+3)} = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)} \Rightarrow 2x+1 = a(x+3) +b(x-1) \ \ \star$
Put $x= 1$ then $a = 3/4$ and $x = -3$ then $b =5/4 $ . I was wondering why this method works correctly because the relation $\star$ works for all values except $1$ and $-3$ . I've tried many examples and strangely this method was correct all the time .